4.11 – Practice Problem 2¶
Saturated Steam table B.6¶
In [7]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/Sat-steam-p-int1.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df.truncate(before = 25, after = 38)
Out[7]:
| Pressure (bar) | Temperature (C) | Volume (l, m3/kg) | Volume (v, m3/kg) | Internal Energy (l, kJ/kg) | Internal Energy (v, kJ/kg) | Enthalpy (l, kJ/kg) | Enthalpy (v, kJ/kg) | |
|---|---|---|---|---|---|---|---|---|
| 25 | 5.5 | 155.46 | 0.001097 | 0.34260 | 655.16 | 2563.9 | 655.76 | 2752.3 |
| 26 | 6.0 | 158.83 | 0.001101 | 0.31558 | 669.72 | 2566.8 | 670.38 | 2756.1 |
| 27 | 6.5 | 161.98 | 0.001104 | 0.29259 | 683.36 | 2569.4 | 684.08 | 2759.6 |
| 28 | 7.0 | 164.95 | 0.001108 | 0.27277 | 696.23 | 2571.8 | 697.00 | 2762.8 |
| 29 | 7.5 | 167.75 | 0.001111 | 0.25551 | 708.40 | 2574.0 | 709.24 | 2765.6 |
| 30 | 8.0 | 170.41 | 0.001115 | 0.24034 | 719.97 | 2576.0 | 720.86 | 2768.3 |
| 31 | 8.5 | 172.94 | 0.001118 | 0.22689 | 731.00 | 2577.9 | 731.95 | 2770.8 |
| 32 | 9.0 | 175.35 | 0.001121 | 0.21489 | 741.55 | 2579.6 | 742.56 | 2773.0 |
| 33 | 9.5 | 177.66 | 0.001124 | 0.20410 | 751.67 | 2581.2 | 752.74 | 2775.1 |
| 34 | 10.0 | 179.88 | 0.001127 | 0.19436 | 761.39 | 2582.7 | 762.52 | 2777.1 |
| 35 | 10.5 | 182.01 | 0.001130 | 0.18552 | 770.75 | 2584.1 | 771.94 | 2778.9 |
| 36 | 11.0 | 184.06 | 0.001133 | 0.17745 | 779.78 | 2585.5 | 781.03 | 2780.6 |
| 37 | 11.5 | 186.04 | 0.001136 | 0.17006 | 788.51 | 2586.7 | 789.82 | 2782.2 |
| 38 | 12.0 | 187.96 | 0.001138 | 0.16326 | 796.96 | 2587.8 | 798.33 | 2783.7 |
Excerpt off steam table of B.7¶
In [8]:
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
df = pd.read_excel('../figures/Module-4/superheated-team-p10.xlsx', sheet_name='Sheet2', index_col=None, na_values=['NA'])
df
Out[8]:
| Temperature (C) | Pressure (bar) | Volume (m3/kg) | Internal Energy (kJ/kg) | Enthalpy (kJ/kg) | Entropy (J/g*K) | |
|---|---|---|---|---|---|---|
| 0 | 50.00 | 10 | 0.001012 | 209.18 | 210.19 | 0.70335 |
| 1 | 100.00 | 10 | 0.001043 | 418.80 | 419.84 | 1.30650 |
| 2 | 150.00 | 10 | 0.001090 | 631.41 | 632.50 | 1.84120 |
| 3 | 179.88 | 10 | 0.001127 | 761.39 | 762.52 | 2.13810 |
| 4 | 179.88 | 10 | 0.194360 | 2582.70 | 2777.10 | 6.58500 |
| 5 | 200.00 | 10 | 0.206020 | 2622.20 | 2828.30 | 6.69550 |
| 6 | 250.00 | 10 | 0.232750 | 2710.40 | 2943.10 | 6.92650 |
| 7 | 300.00 | 10 | 0.257990 | 2793.60 | 3051.60 | 7.12460 |
| 8 | 350.00 | 10 | 0.282500 | 2875.70 | 3158.20 | 7.30290 |
Modified question of 7.35¶
Problem Statement
A turbine discharges 200 kg/h of saturated steam at 10.0 bar absolute. It is desired to generate steam at 300 C and 10.0 bar by mixing the turbine discharge with a second stream of superheated steam of 350 C and 10.0 bar.
- If 300 kg/h of the product steam is to be generated, how much heat must be added to the mixer?
- If instead the mixing is carried out adiabatically, at what rate is the product steam generated?
Preamble: This question is relatively basic and engages you on how to use the steam tables. Heat is covered in-depth in the next module but for now, you can consider the question as, how much more enthalpy is needed to be added to the mixer.
- Draw a diagram of the process taking place and write the relevant equation:
\[\dot{m_1}H_1 + \dot{m_2}H_2 = \dot{m_3}H_3\]
\[H = Q -W \space or \space H_{total} = \sum {H}\]
- List simplifications and assumptions
- As mentioned above, treat heat as enthalpy.
- If you have read into module 5; then assume work is negligable.
- List known values
- Input Stream 1:
- \(\dot{m_1} = 200\frac{kg}{hr}\)
- \(H_1 = 762.5 \frac{kj}{kg}\)
- Input Stream 2:
- \(\dot{m_2} = ?\)
- \(H_2 = 3051.6 \frac{kj}{kg}\) From steam table B.6
- Output Stream 3:
- \(\dot{m_3} = ?\)
- \(H_3 = 3158.2 \frac{kj}{kg}\) From steam table B.7
- Solve the equations. In part i) we are given the total mass flow rate \(\dot{m_3}\) So we can solve for \(\dot{m_2}\) using conservation of mass.
\[300 = 200 + \dot {m_2}\]
Using this value we plug into the first half of the energy balance
\[200\frac{kg}{hr} \cdot 762.5 \frac{kj}{kg} + 100\frac{kg}{hr} \cdot 3051.6 \frac{kj}{kg} = 457660 \frac{kj}{hr}\]
Since we want the exit stream at 350 C we need to check to see if the enthalpy we obtained corresponds with the desired enthalpy in the exit stream.
\[\dot{m_3}\cdot H_3 = 300\frac{kg}{hr} \cdot 3158.2 \frac{kj}{kg} = 947460 \frac{kj}{hr}\]
It Doesn’t correspond. There is missing energy that needs to be added. This is simple addition:
\[Energy\space Needed = 947460 \frac{kj}{hr} - 457660 \frac{kj}{hr} = 489800 \frac {kj}{hr}\]
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